2 Microelettronica – Circuiti integrati analogici 2/edRichard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Chapter 13Frequency Response
Microelectronic Circuit Design
Richard C. Jaeger
Travis N. Blalock
2 Microelettronica – Circuiti integrati analogici 2/edRichard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Chapter Goals• Review transfer function analysis and dominant-pole
approximations of amplifier transfer functions.• Learn partition of ac circuits into low and high-frequency
equivalents.• Learn short-circuit and open-circuit time constant methods to
estimate upper and lower cutoff frequencies.• Develop bipolar and MOS small-signal models with device
capacitances.• Study unity-gain bandwidth product limitations of BJTs and
MOSFETs.• Develop expressions for upper cutoff frequency of inverting, non-
inverting and follower configurations.• Explore gain-bandwidth product limitations of single and multiple
transistor circuits.
2 Microelettronica – Circuiti integrati analogici 2/edRichard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Chapter Goals (contd.)• Understand Miller effect and design of op amp frequency
compensation.
• Develop relationship between op amp unity-gain frequency and slew rate.
• Understand use of tuned circuits to design high-Q band-pass amplifiers.
• Understand concept of mixing and explore basic mixer circuits.
• Study application of Gilbert multiplier as balanced modulator and mixer.
2 Microelettronica – Circuiti integrati analogici 2/edRichard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Transfer Function Analysis
)()(
...2210
...2210
)()()(
sHFsLFmid
A
nsnbsbsbb
msmasasaa
sDsNsvA
Amid is midband gain between upper and lower cutoff frequencies.
HPl
sHP
sHP
s
HZl
sHZ
sHZ
s
sHF
LPk
sLPsL
Ps
LZk
sLZsL
ZssLF
1...
2
1
1
1
1...
2
1
1
1
)(
...21
...21)(
1)( jHF HPi
HZi,for ,i =1…l
)()( sLFmid
AsLA
1)( jLF for ,j =1…k)()( sHF
midAsHA
LPj
LZj,
2 Microelettronica – Circuiti integrati analogici 2/edRichard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Low-Frequency Response
2
2
)(
PLP
sssLF
Pole P2 is called the dominant low-frequency pole (> all other poles) and zeros are at frequencies low enough to not affect L.
If there is no dominant pole at low frequencies, poles and zeros interact to determine L.
21
21)()(
PsPsZsZs
midAsLF
midAsLA
For s=j, at L, 2)( mid
A
LjA
2
222
12
22
221
2
21
PLPL
ZLZL
4
22
21
2
22
211
4
22
21
2
22
211
21
L
PP
L
PP
L
ZZ
L
ZZ
Pole L > all other pole and zero frequencies
In general, for n poles and n zeros,
22
221
222
21 ZZPPL
n Znn PnL
222
2 Microelettronica – Circuiti integrati analogici 2/edRichard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Transfer Function Analysis and Dominant Pole Approximation Example• Problem: Find midband gain, FL(s) and fL for
• Analysis: Rearranging the given transfer function to get it in standard form,
Now,
and
Zeros are at s=0 and s =-100. Poles are at s= -10, s=-1000
All pole and zero frequencies are low and separated by at least a decade. Dominant pole is at =1000 and fL =1000/2= 159 Hz. For frequencies > a few rad/s:
100011.0
11002000)(
ss
sssLA
100010
100200)( ss
sssLA
200
)()(
midA
sLFmid
AsLA
)1000)(10()100()(
ss
sssLF
Hz158210020(2210002102
1 Lf
1000200)(
sssLA
2 Microelettronica – Circuiti integrati analogici 2/edRichard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
High-Frequency Response
3
)3
/(1)(
PH
Ps
ssLF
Pole P3 is called the dominant high-frequency pole (< all other poles).
If there is no dominant pole at low frequencies, poles and zeros interact to determine H.
)2/(1)1/(1
)2/(1)1/(1
)()(
PsPsZsZs
midA
sHFmid
AsHA
For s=j, at H, 2)( mid
A
HjA
)2
2/2(1)2
1/2(1
)22
/2(1)21
/2(1
2
1
PHPH
ZHZH
22
21
4
22
2
21
21
22
21
4
22
2
21
21
21
PP
H
P
H
P
H
ZZ
H
Z
H
Z
H
Pole H < all other pole and zero frequencies
In general,
22
22
1
22
2
12
1
11
ZZPP
H
n Znn Pn
H
212
21
1
2 Microelettronica – Circuiti integrati analogici 2/edRichard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Direct Determination of Low-Frequency Poles and Zeros: C-S Amplifier
(s)gsV
)3(2
1)3(
33
)2
/1(3
(s)gsV3(s)oI(s)oV
RDRCs
sD
RRmg
RRsCR
DR
mgR
(s)gV
)/1(3
1
)3
/1(sV-gV(s)gsV
SRmgC
s
SRCs
IRGRGR
DRRmg
midA
(s)LFmid
A(s)vA
)3(
(s)i
V
(s)oV
(s)i
V1)(
1
1(s)gV
GR
IRCs
GRCs
2 Microelettronica – Circuiti integrati analogici 2/edRichard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Direct Determination of Low-Frequency Poles and Zeros: C-S Amplifier (contd.)
The three zero locations are: s = 0, 0, -1/(RS C3).The three pole locations are:
Each independent capacitor in the circuit contributes one pole and one zero. Series capacitors C1 and C2 contribute the two zeros at s=0 (dc), blocking propagation of dc signals through the amplifier. Third zero due to parallel combination of C3 and RS occurs at frequency where signal current propagation through MOSFET is blocked (output voltage is zero).
)3
(2
1
)/1(3
1)(
1
1
)3
/1(2
(s)
RD
RCs
SRmgC
s
GR
IRC
s
SRCss
LF
)3(2
1,)/1(3
1,)(1
1sRDRC
SRmgCGRIRC
2 Microelettronica – Circuiti integrati analogici 2/edRichard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Short-Circuit Time Constant Method to Determine L • Lower cutoff frequency for a
network with n coupling and bypass capacitors is given by:
where RiS is resistance at terminals of ith capacitor Ci with all other capacitors replaced by short circuits. Product RiS Ci is short-circuit time constant associated with Ci.
n
i iC
iSRL
1
1
Midband gain and upper and lower cutoff frequencies that define bandwidth of amplifier are of more interest than complete transfer function.
2 Microelettronica – Circuiti integrati analogici 2/edRichard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Estimate of L for C-E AmplifierUsing SCTC method, for C1,
For C2,
For C3,
)(2)(1 rBRRRCEinBRIRSR
CRRorCRRRCE
outCRRSR
3
)(3)(32
1
)(
13
o
BR
IRr
ER
o
thRr
ERRCCoutERSR
3
1
1
i iC
iSRL
2 Microelettronica – Circuiti integrati analogici 2/edRichard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Estimate of L for C-S AmplifierUsing SCTC method,
For C1,
For C2,
For C3,
GRSRRCSinGRIRSR )(1
DRRorDRRRCS
outDRRSR
3
)(3)(32
mgSRRCGoutSRSR 1
3
2 Microelettronica – Circuiti integrati analogici 2/edRichard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Estimate of L for C-B Amplifier
Using SCTC method,
For C1,
For C2,
)1()(1mgERIRRCB
inERIRSR
CRRRCBoutCRRSR 3)(32
2 Microelettronica – Circuiti integrati analogici 2/edRichard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Estimate of L for C-G Amplifier
Using SCTC method,
For C1,
For C2,
)1()(1mgSRIRRCG
inSRIRSR
DRRRCGoutDRRSR 3)(32
2 Microelettronica – Circuiti integrati analogici 2/edRichard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Estimate of L for C-C Amplifier
Using SCTC method,
For C1,
For C2,
31
)(1
RE
RorB
RIR
RCCinBRIRSR
13)(32o
thRr
ERRRCC
outERRSR
2 Microelettronica – Circuiti integrati analogici 2/edRichard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Estimate of L for C-D Amplifier
Using SCTC method,
For C1,
For C2,
GRIRRCDinGRIRSR )(1
mgSRRRCDoutSRRSR 1
332
2 Microelettronica – Circuiti integrati analogici 2/edRichard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Frequency-dependent Hybrid-Pi Model for BJT
Capacitance between base and collector terminals is:
Co is total collector-base junction capacitance at zero bias, jc is its built-in potential.
)/(1jcCB
VoC
C
Capacitance between base and emitter terminals is:
F is forward transit-time of the BJT. Cappears in parallel with r. As frequency increases, for a given input signal current, impedance of C reduces vbe and thus the current in the controlled source at transistor output.
FmgC
2 Microelettronica – Circuiti integrati analogici 2/edRichard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Unity-gain Frequency of BJT
1)(
1
(s)b
I
(s)cI)(
1)((s)
b)I(
(s)be
)V((s)cI
rCCsmg
sC
os
rCCs
rsCmg
sCmg
The right-half plane transmission zero Z = + gm/C occurring at high frequency can be neglected.
= 1/ r(C + C ) is the beta-cutoff frequency
where
and fT = T /2 is the unity gain bandwidth product. Above BJT has no current gain.
1)/(1)()(
so
rCCsos
s
Ts
os)(
CCmg
rCCo
oT
)(
2 Microelettronica – Circuiti integrati analogici 2/edRichard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Unity-gain Frequency of BJT (contd.)
Current gain is o = gmr at low frequencies and has single pole roll-off at frequencies > f, crossing through unity gain at T. Magnitude of current gain is 3 dB below its low-frequency value at f.
C
T
CI
C
T
mgC
40
2 Microelettronica – Circuiti integrati analogici 2/edRichard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
High-frequency Model of MOSFET
)/(11
(s)gI
(s)d
I)(
)(
)((s)
bI
(s)gs)V((s)d
I
GDC
GSC
T
ssTs
GDC
GSCs
GDsCmg
GDsCmg
CCmg
rCCo
oT
)(
22
3
"(2/3)
"
L
TNV
GSVn
WLoxC
TNV
GSV
L
WoxCn
Tf
2 Microelettronica – Circuiti integrati analogici 2/edRichard C. Jaeger, Travis N. Blalock
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Limitations of High-frequency Models
• Above 0.3 fT, behavior of simple pi-models begins to deviate significantly from the actual device.
• Also, T depends on operating current as shown and is not constant as assumed.
• For given BJT, a collector current ICM exists that yield maximum fTmax.
• For FET in saturation, CGS and CGD are independent of Q-point current, so
DImgT
2 Microelettronica – Circuiti integrati analogici 2/edRichard C. Jaeger, Travis N. Blalock
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Effect of Base Resistance on Midband Amplifiers
Base current enters the BJT through external base contact and traverses a high resistance region before entering active area. rx models voltage drop between base contact and active area of the BJT.
To account for base resistance rx is absorbed into equivalent pi model and can be used to transform expressions for C-E, C-C and C-B amplifiers.
xrro
xrrr
mgmg
mgxrr
rmgmg
'
bev'
bevvi
xrrr ' oo '
2 Microelettronica – Circuiti integrati analogici 2/edRichard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Direct High-Frequency Analysis: C-E Amplifier
The small-signal model can be simplified by using Norton source transformation.
kΩ3.4kΩ1003 C
RRLR kΩ10kΩ3021 RRBR
BRIRBR
ivth
v BRIR
BRIRth
R
xrthR th
vsi )( xrth
Rror
2 Microelettronica – Circuiti integrati analogici 2/edRichard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Direct High-Frequency Analysis: C-E Amplifier (Pole Determination)
From nodal equations for the circuit in frequency domain,
High-frequency response is given by 2 poles, one finite zero and one zero at infinity. Finite right-half plane zero, Z = + gm/C > T can easily be neglected.
For a polynomial s2+sA1+A0 with roots a and b, a =A1 and b=A0/A1.
ogLgogL
CggL
gCL
gCs
LCC
LCCCs
mgsC
2
)-((s)sI(s)2V
LCC
mg
LCC
LCC
mgP
TCorA
A
P
orL
R
LCorL
R
LRmgCCTC
)/(12
1
1
01
1
Smallest root that gives first pole limits frequency response and determines H. Second pole is important in frequency compensation as it can degrade phase margin of feedback amplifiers.
2 Microelettronica – Circuiti integrati analogici 2/edRichard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Direct High-Frequency Analysis: C-E Amplifier (Overall Transfer Function)
)1/(1(s)th
V(s)oV
)(
)1
/(1
(s)th
V-(s)oV
)2
/(1)1
/(1
)/(1)(
(s)th
V(s)oV
)2
/(1)1
/(1
)-((s)th
V(s)oV
Psmid
As
vthA
Ps
orL
Rmg
xrthR
Ps
Psog
Lg
Zs
orLRmgxrth
R
Ps
Psog
Lg
mgsC
xrthR
rxrthR
LRo
midA
TCorP
1
1
Dominant pole model at high frequencies for C-E amplifier is as shown.
2 Microelettronica – Circuiti integrati analogici 2/edRichard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Direct High-Frequency Analysis: C-E Amplifier (Example)• Problem: Find midband gain, poles, zeros and fL.
• Given data: Q-point= ( 1.60 mA, 3.00V), fT =500 MHz, o =100, C
=0.5 pF, rx =250CL
• Analysis: gm =40IC =40(0.0016) =64 mS, r = o/gm =1.56 k.pF9.19
2 C
Tfmg
C
kΩ12.4kΩ3.4kΩ1003 C
RRLR
882ΩkΩ1kΩ5.7 I
RBRth
R
656)( xrthRror
pF1561
orL
R
LCorL
R
LRmgCCTC
MHz56.12
11
TCorPf
L
RmgormgCmg
CL
RP1111
2
153
rxrthR
LRo
vthA
GHz4.202
C
mgZfMHz603
22
2
P
Pf
Overall gain is reduced to -135 as vth =0.882vs.
2 Microelettronica – Circuiti integrati analogici 2/edRichard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Gain-Bandwidth Product Limitations of C-E Amplifier
• If Rth is reduced to zero in order to increase bandwidth, then ro would not be zero but would be limited to approximately rx.
• If Rth = 0, rx <<r so that rx = ro and
TCorrxrth
RL
RoHvA
1GBW
)( LRmgCTC
Cxr1GBW
2 Microelettronica – Circuiti integrati analogici 2/edRichard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
High-Frequency Analysis: C-S Amplifier
GRIR
thR
3RDRLR
GRIRGR
ivth
v
TC
thRP
11
th
RL
R
LRmgGDCGSCTC 1
L
Rmgth
RmgGS
Cmg
GDC
LRP
11112
GDC
mgZ
2 Microelettronica – Circuiti integrati analogici 2/edRichard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Miller Multiplication
)1()(
1V
)(sI)(Y
)(1V)(oV
AsCs
ss
sAs
Total input capacitance = C(1+A) because total voltage across C is vc = vi(1+A) due to inverting voltage gain of amplifier.
)(oV)(1
V)(sI sssCs
For the C-E amplifier,
)1()1( LRmgCCACCTC
2 Microelettronica – Circuiti integrati analogici 2/edRichard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Miller Integrator
Assuming zero current in input terminal of amplifier,
inVoV
)oVin
V(inV
1V
A
sCR
osoA
ARCs
AA
RCsvA
)1(11
1
1V
oV)(
)1(1
ARCo where
For frequencies >> o, assuming A>>1,
which is the transfer function of an integrator.
sRCsoA
svA 1)(
2 Microelettronica – Circuiti integrati analogici 2/edRichard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Open-Circuit Time Constant Method to Determine H
At high frequencies, impedances of coupling and bypass capacitors are small enough to be considered short circuits. Open-circuit time constants associated with impedances of device capacitances are considered instead.
m
ii
Cio
RH
1
1
where Rio is resistance at terminals of ith capacitor Ci with all other capacitors open-circuited.
For a C-E amplifier, assuming CL =0oroR
)1(xixv
orL
R
LRmgoroR
TCorCoRCoRH
11
2 Microelettronica – Circuiti integrati analogici 2/edRichard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Gain-Bandwidth Trade-off Using Emitter Resistor
ER
LR
ERorxrth
RL
Romid
A
)1(
xrthRr for and
gain decreases as emitter resistance increases and bandwidth of stage will correspondingly increase.
To find bandwidth using OCTC method:
1ERmg
ERmg
ERxrth
ReqR
1xixv
ERmg
ERxrth
R
eqRroR
1
2 Microelettronica – Circuiti integrati analogici 2/edRichard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Gain-Bandwidth Trade-off Using Emitter Resistor (contd.)
Test source ix is first split into two equivalent sources and then superposition is used to find vx =(vb - vc).
Assuming that o >>1 and
xrthR
LR
ERmgLRmg
xrthRoR
11)(
xixv
ERorxrthR )1(
xrthR
LR
ERmgL
RmgC
xrthR
ER
ERmg
Cxrth
R
H
111
1)(
1
2 Microelettronica – Circuiti integrati analogici 2/edRichard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Dominant Pole for C-B Amplifier
IRER
thR
3RCRLR
thRmg
xrthR
thRmg
xrthR
rroR
11xixv
Using split-source transformation Assuming that o >>1 and rx << r
LRthRmg
LRmgxroR
11
xicvbv
LRC
thRmgL
RmgC
xrth
R
thRmg
Cxr
H
111
1
1
Neglecting first term of order of 1/ T and since last term is dominant.
LRCH
1
2 Microelettronica – Circuiti integrati analogici 2/edRichard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Dominant Pole for C-G Amplifier
IRR
thR 4
3RDRLR
mgth
GthRmg
thR
GSoR
11
LRGDoR
LR
GDC
LR
GDC
mgth
GGS
CH11
2 Microelettronica – Circuiti integrati analogici 2/edRichard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Dominant Pole for C-C Amplifier
IRBR
thR
3RERLR
LRmg
LRxrth
R
LRmg
LRxrth
RrroR
11xixv
)(
)1()()(
xrthR
LRorxrth
RRCCinxrth
RoR
Cxrth
R
LRmg
CL
RxrthR
H)(
1)(
1
A better estimate is obtained if we set RL =0 in expression for Ro.
C
LRmg
Cxrth
RH
1)(
1
xrCH
1)1(GBW
2 Microelettronica – Circuiti integrati analogici 2/edRichard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Dominant Pole for C-D Amplifier
Substituting ras infinite and rx as zero in expression for emitter follower,
IRGR
thR
3RSRLR
GDC
LRmg
GSC
thRth
RGD
CGS
C
LRmg
thRH
1
1
1
1
2 Microelettronica – Circuiti integrati analogici 2/edRichard C. Jaeger, Travis N. Blalock
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Frequency Response: Differential Amplifier
CEE is total capacitance at emitter node of the differential pair.
Differential mode half-circuit is similar to a C-E stage. Bandwidth is determined by the product. As emitter is a virtual ground, CEE has no effect on differential-mode signals.
For common-mode signals, at very low frequencies,
Transmission zero due to CEE is
TCor
12
)0( EE
RC
RccA
EER
EECZs 1
2 Microelettronica – Circuiti integrati analogici 2/edRichard C. Jaeger, Travis N. Blalock
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Frequency Response: Differential Amplifier (contd.)
Common-mode half-circuit is similar to a C-E stage with emitter resistor 2REE. OCTC for C and C is similar to the C-E stage. OCTC for CEE/2 is:
mgo
xrrEEREEOR 1
12
mgEE
C
xrC
R
EERmgC
RmgC
xrEE
R
EERmg
Cxr
P
2211
21
21
1
As REE is usually designed to be large,
)(1
)(2
1
xrCRC
xrCRC
mgEE
CCP
2 Microelettronica – Circuiti integrati analogici 2/edRichard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Frequency Response: Common-Collector/ Common-Base Cascade
REE is assumed to be large and neglected.
Sum of the OCTC of Q1 is:
1
111
111
mgo
xrrRCC
out
2
112
222
mgo
xrrRCB
in
12
111
2
11
1
11
CC
xrC
mgm
g
C
xr
2 Microelettronica – Circuiti integrati analogici 2/edRichard C. Jaeger, Travis N. Blalock
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Frequency Response: Common-Collector/ Common-Base Cascade (contd.)
Sum of the OCTC of Q2 is:
Combining the OCTC for Q1 and Q2, and assuming that transistors are matched,
2
21
222
22
1
12
11
2
1
12
1
22
xr
CR
CRmg
CC
xrCRC
mgm
g
CRmg
C
mgm
g
C
xr
xrC
RC
RmgCCxr
H
22
1
2 Microelettronica – Circuiti integrati analogici 2/edRichard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Frequency Response: Cascode Amplifier
OCTC of Q1 with load resistor 1/ gm2 :
As IC2 = IC1, gm2 = gm1, gain of first stage is unity. Assuming gm2 ro1 >>1,
OCTC of Q1, a C-B stage for ro1 >> RL and f>>1:
Assuming matched devices,
12
1
2
11111111111
or
mg
mg
mg
CCorTCorCoRCoR
12
1111 CCorTCor
2)2(
2
22222
CLRxr
mg
CCoRCoR
C
LRxrCC
orH
21
1
2 Microelettronica – Circuiti integrati analogici 2/edRichard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Frequency Response: MOS Current Mirror
1
1
mgor 2orLR
21 GSCGSCC 2GDCC
222
1
222
1
2
11
or
GDC
or
GDC
mg
GSCP
For matched transistors,
2 Microelettronica – Circuiti integrati analogici 2/edRichard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Frequency Response: Multistage Amplifier• Problem:Use open-circuit and short-circuit time constant methods to
estimate upper and lower cutoff frequencies and bandwidth.• Approach: Coupling and bypass capacitors determine low-frequency
response, device capacitances affect high-frequency response.
At high frequencies, ac model for multi-stageamplifier is as shown.
2 Microelettronica – Circuiti integrati analogici 2/edRichard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Frequency Response: Multistage Amplifier (Estimate of L)
SCTC for each of the six independent coupling and bypass capacitors has to be determined.
1.01MΩ
1MΩkΩ10)1
(1
in
RGRIRSR
7.66S01.0
1200
1
112
mgSRSR
kΩ69.22211
22113
rBR
or
DR
inR
BR
OR
DRSR
5711122 o
rD
RBRth
R
4.19
12
2224
o
rth
R
ERSR
kΩ4.18
)3
)(13
(3322
33225
LR
ER
or
BR
or
CR
inR
BR
OR
CRSR
k99.32233 o
rC
RBRth
R
311
13
3336
o
rth
R
ERLRSR
Hz5302
rad/s33001
1
LLf
n
i iC
iSRL
2 Microelettronica – Circuiti integrati analogici 2/edRichard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Frequency Response: Multistage Amplifier (Estimate of H)
OCTC for each of the two capacitors associated with each transistor has to be determined.For M1,
For Q2,
4782121 rIRLR
s71007.1
111
1111
thR
LR
LR
mg
GDC
GSC
thRTC
thR
5701122 o
rIRth
R
610)22
(22 xr
thRror
kΩ54.3
)3
)(13
(323
3232
LR
ER
or
IR
inR
IRLR
s71074.1
2
222
122222
or
LR
LR
mgCCorTCor
For Q3, k99.32233 o
rIRth
R
s81051.1
3)33(3
31
)33
(
3333
Cxrth
RC
EER
mg
xr
thR
CORCOR
kHz5382
rad/s61038.3
1
1
HHf
m
ii
Cio
RH
2 Microelettronica – Circuiti integrati analogici 2/edRichard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Single-pole Op Amp Compensation• Frequency compensation forces overall amplifier to have a
single-pole frequency response by connecting compensation capacitor around second gain stage of the basic op amp.
Bs
T
Bs
BoAsvA
)(
2 Microelettronica – Circuiti integrati analogici 2/edRichard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Three-stage MOS Op Amp AnalysisInput stage is modeled by its Norton equivalent- current source Gmvdm and output resistance Ro. Second stage has gain of gm5ro5= f5 and follower output stage is a unity-gain buffer.Vo(s) = Vb(s) = - Av2Va(s)
For large Av2
Bs
BoA
Bs
T
v2A
CCosR1
v2AoRmG
s
sv2
A
s
ssvA
)1()(dm
V
)(aV-
)(dm
V
)(oV)(
)1(v2
AC
CoR
1B
)1(v2
AC
Cv2
AmG
T
CC
mGT
2 Microelettronica – Circuiti integrati analogici 2/edRichard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Transmission Zeros in FET Op Amps
Incorporating the zero determined by gm5 in the analysis,
This zero can’t be neglected due to low ratio of transconductances of M2 and M5. Zero can be canceled by addition of RZ =1/ gm5.
)1/(1
)/(1)55()(
PsZs
ormgsvth
A
oR
or
fGDCCCGSCTC 55
1)5(5
TCoRP1
1
2
5
5
5
mg
mg
TGD
CC
Cm
g
Z
CC
ZR
mg
Z
)5
/1(
2 Microelettronica – Circuiti integrati analogici 2/edRichard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Bipolar Amplifier Compensation• Bipolar op amp can be
compensated in the same manner as a MOS amplifier
• Transmission zero occurs at too high a frequency to affect the response due to higher transconductance of BJT that FET for given operating current.
• Unity gain frequency is given by:
2
55
CIC
I
TC
Cm
g
Z
CC
CI
CC
CI
CCm
g
T1
202
402
2 Microelettronica – Circuiti integrati analogici 2/edRichard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Slew rate of Op Amp• Slew-rate limiting is caused by
limited current available to charge/discharge internal capacitors. For very large Av2, amplifier behaves like an integrator:
• For CMOS amplifier,
• For bipolar amplifier,
dt
todvCC
dt
tB
dv
CCCI)()(
1
1
/1
max
)(SR
ImGT
CCI
dt
todv
21
1/
SRnKI
TImGT
201
/SR T
ImGT
2 Microelettronica – Circuiti integrati analogici 2/edRichard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Tuned Amplifiers
• Amplifiers with narrow bandwidth are often required in RF applications to be able to select one signal from a large number of signals.
• Frequencies of interest > unity gain frequency of op amps, so active RC filters can’t be used.
• These amplifiers have high Q (fH and fL close together relative to center frequency)
• These applications use resonant RLC circuits to form frequency selective tuned amplifiers.
2 Microelettronica – Circuiti integrati analogici 2/edRichard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Single-Tuned Amplifiers• RLC network selects the
frequency, parallel combination of RD, R3 and ro set the Q and bandwidth.
• Neglecting right-half plane zero,
3
)/1()()(i
V
)(oV)(
GDGogPG
sLGD
CCsP
Gmg
GDsC
s
ssvA
22)(
oQoss
Qos
midAsvA
)(1
GDCCLo
Lo
PR
GDCCPRoQ
)(
2 Microelettronica – Circuiti integrati analogici 2/edRichard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Single-Tuned Amplifiers (contd.)
• At center frequency, s = jo, Av = Amid.
)3
( RD
RormgPRmgmid
A
PR
Lo
GDCC
PRQ
o2
)(1BW
2 Microelettronica – Circuiti integrati analogici 2/edRichard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Use of tapped Inductor- Auto Transformer
CGD and ro can often be small enough to degrade characteristics of the tuned amplifier. Inductor can be made to work as an auto transformer to solve this problem.
These results can be used to transform the resonant circuit and higher Q can be obtained and center frequency doesn’t shift significantly due to changes in CGD.
Similar solution can be used if tuned circuit is placed at amplifier input instead of output
)(sI)(1V2
/)(sI)(1nV
)(2I)(oV
ss
nns
sss )(2)( spZnssZ
2 Microelettronica – Circuiti integrati analogici 2/edRichard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Multiple Tuned Circuits• Tuned circuits can be placed at both input
and output to tailor frequency response.• Radio-frequency choke(an open circuit at
operating frequency) is used for biasing. • Synchronous tuning uses two circuits
tuned to same center frequency for high Q.
• Stagger tuning uses two circuits tuned to slightly different center frequencies to realize broader band amplifiers.
• Cascode stage is used to provide isolation between the two tuned circuits and eliminate feedback path between them due to Miller multiplication.
1/121BWnBW n
2 Microelettronica – Circuiti integrati analogici 2/edRichard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Mixers: Conversion Gain
• Amplifiers discussed so far have always been assumed to be linear and gain expressions involve input and output signals at same frequency.
• Mixers are nonlinear devices, output signal frequency is different from input signal frequency.
• A mixer’s conversion gain is the ratio of phasor representation of output signal to that of input signals, the fact that the two signals are at different frequencies is ignored.
2 Microelettronica – Circuiti integrati analogici 2/edRichard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Single-Balanced Mixer• Eliminates one of the two input
signals from the output.
• No signal energy appears at1 , but 2 appears in output spectrum, so circuit is single-balanced.
• Up-conversion uses component (2-1) and down-conversion uses (2+1) component.
tnoddn n
tv
tIEEIEEi
2sin4)(2
1sin1
tnC
RItnC
RI
tnC
REE
I
oddn ntoV
)12
cos(2
1)12
cos(2
1
2sin
4)(
2 Microelettronica – Circuiti integrati analogici 2/edRichard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Double-Balanced Mixer/ Modulator: Gilbert Multiplier
• Double-balanced mixers don’t contain spectral components at either of the two input frequencies.
• Modulator applications give double sideband suppressed carrier output signal. Amplitude-modulated signal can also be obtained if
tmRmV
BBICi sin
121 tmR
mVBBICi sin
122
tmcntmcn
oddn nRCR
mVtoV )cos()cos(4
1)(
tmcnMtmcnMtcn
oddn nRCR
mVtov )cos(2
)cos(2
sin4
1)(
)sin1(1 tmMmVv